Parabolas+B-1

=Parabolas, Completing the Square, and Averaging the X-intercepts= 

=Parabolas=

A parabola's shape is a U. To make parabola you can Complete the Square or you could Average the X-intercepts. The two forms for parabolas are the following:


 * Graphing form: ** y=(x-h)²+k


 * Standard form: **y=ax²+bx+c.

=Completing the Square=

Completing the square changes the equation of the parabola from standard form to graphing form.

You want to get the equation from this: to this: The **'-h'** and '**k'** variables are the vertex of the graph of the equation. The variables -h and k are for the vertex. -h is for the x value of the vertex and k is for the y value of the vertex. The variable **'a'** determines the width of the graph of the equation.

__Example #1__

math y=x^2+3x+5 math


 * (Step 0)** First make sure that there is 1 in front of the x².
 * (Step 1)** Then take half of b which is 3 then move c which is 5 down. So this is what it looks like:

math y=(x + 1.5)^2 +5 math


 * (Step 2)** Next thing you do is multiply b which is 1.5, to the second power and subtract it by c which is 5. The equation now looks like:

math y= (x + 1.5)^2 + 5 - 2.25 math

The resulting equation is now:

math y= (x + 1.5)^2 + 2.75 math

And the resulting vertex is: (-1.5, 2.75)

__Example #2__

 =Averaging the x-intercepts=

This is another method that can be used for changing from the standard form to the graphing form. The value of "a" in the graphing form of the equations comes from the value of "a" in the standard form that you were given. Now you have to figure out the //**vertex**// by using the average of the x- intercepts.

Example: Change this equation to graphing form: y= x²- 6x + 5

The x intercepts of this parabola would be (1,0) and (5,0) the vertex would then be whats in the middle and whatever it is the things would be (h,k) so when the vertex is 3 but you don't know the k you can just plug that in.

y= a(x-3)² + k

'a' would then be 1 which comes from the coefficient of x² in the original equation.

y=1 (x-3)²+k

So then we need to find out what K is so what we do is change y to 0 and then x to 1. The new question would be:

0=1(1-3)²+k 0= (-2)²+k 0=4+k

you then subtract 4 from both sides.

-4=k

So now since you know K you can plug it in and have your problem.

y= 1(x-3)² - 4

Check Your Understanding
When I say page numbers you can refer to the Algebra 2 textbook in the classroom or on the hotmath website (hotmath.com)

COMPLETING THE SQUARE pg 529

1. y=x^2 -6x +9

2. y=x^2+3

6. y= x^2- 1/3x

7. y= 3x^2-6x + 1

8. y=5x^2 + 20x- 16

AVERAGING THE X INTERCEPTS pg. 278

CT-6 Find the equation o the parabola that passes through the points (0,0), (3,9) and (6,0). 